What is the derivative of \sqrt[3]{7x^6+8x^5} ?

The derivative of \sqrt[3]{7x^6+8x^5} is (42x^5+40x^4)/(3(7x^6+8x^5)^{2/3)}

What is the first derivative of \sqrt[3]{7x^6+8x^5} ?

The first derivative of \sqrt[3]{7x^6+8x^5} is (42x^5+40x^4)/(3(7x^6+8x^5)^{2/3)}

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derivative of

3 7 x^{6} + 8 x^{5}

\frac{42 x ^{5} + 40 x ^{4}}{3 ( 7 x ^{6} + 8 x ^{5} ) ^{\frac{2}{3}}}

Solution steps

\frac{d}{d x} (3 7 x^{6} + 8 x^{5})

Apply the chain rule:

\frac{1}{3 ( 7 x ^{6} + 8 x ^{5} ) ^{\frac{2}{3}}} \frac{d}{d x} (7 x^{6} + 8 x^{5})

= \frac{1}{3 ( 7 x ^{6} + 8 x ^{5} ) ^{\frac{2}{3}}} \frac{d}{d x} (7 x^{6} + 8 x^{5})

\frac{d}{d x} (7 x^{6} + 8 x^{5}) = 42 x^{5} + 40 x^{4}

= \frac{1}{3 ( 7 x ^{6} + 8 x ^{5} ) ^{\frac{2}{3}}} (42 x^{5} + 40 x^{4})

Simplify

\frac{1}{3 ( 7 x ^{6} + 8 x ^{5} ) ^{\frac{2}{3}}} (42 x^{5} + 40 x^{4}) : \frac{42 x ^{5} + 40 x ^{4}}{3 ( 7 x ^{6} + 8 x ^{5} ) ^{\frac{2}{3}}}

= \frac{42 x ^{5} + 40 x ^{4}}{3 ( 7 x ^{6} + 8 x ^{5} ) ^{\frac{2}{3}}}

sin (x) sin (y) + y^{^{'}} cos (y) = 0

\frac{d}{d x} (3 x^{2} - 6 x + 7)

\frac{x + 1}{x ^{2} + 1}

\frac{d}{d x} (x e^{x^{6}})

2 (12) (18) + 2 (9)

What is the derivative of \sqrt[3]{7x^6+8x^5} ?
The derivative of \sqrt[3]{7x^6+8x^5} is (42x^5+40x^4)/(3(7x^6+8x^5)^{2/3)}
What is the first derivative of \sqrt[3]{7x^6+8x^5} ?
The first derivative of \sqrt[3]{7x^6+8x^5} is (42x^5+40x^4)/(3(7x^6+8x^5)^{2/3)}